Water of mass `m_(2) = 1 kg` is contained in a copper calorimeter of mass `m_(1) = 1 kg`. Their common temperature `t = 10^(@)C`. Now a piece of ice of mass `m_(2) = 2 kg` and temperature is `-11^(@)C` dropped into the calorimeter. Neglecting any heat loss, the final temperature of system is. [specific heat of copper `=0.1 Kcal//kg^(@)C`, specific heat of water `= 1 Kcal//kg^(@)C`, specific heat of ice `= 0.5 Kcal//kg^(@)C`, latent heat of fusion of ice `= 78.7 Kcal//kg`]

Options

A. `0^(@)C`
B. `4^(@)C`
C. `-4^(@)C`
D. `-2^(@)C`

Explanation

Loss in heat from calorimeter + water as temperature changes from `10^(@)C` to `0^(@)C`
`=m_(1)C_(1)10+m_(2)C_(2)10=1xx1xx10+1xx0.1xx10=11` kcal
gain in heat of ice as its temperature changes from `-11^(@)C` to `0^(@)C`
`=m_(3)C_(3)xx11=2xx0.5xx11=11` kcal
Hence ice and water will coexist at `0^(@)C` without any phase change.